NDA Mathematics Question (2026) — Solution

Step-by-step solution

The total number of 4-digit numbers that can be formed using the digits 0, 1, 2, 3, 4 without repetition is calculated by filling the thousands place in 4 ways (excluding 0) and the remaining 3 places in ^ 4 P_ 3 = 24 ways. Total numbers = 4 24 = 96 For a number to be divisible by 6, it must be divisible by both 2 and 3. For divisibility by 3, the sum of the digits must be a multiple of 3. The sum of all five digits is 0 + 1 + 2 + 3 + 4 = 10. To choose 4 digits whose sum is a multiple of 3, the excluded digit must be 1 or 4. Case 1: The excluded digit is 1. The selected digits are 0, 2, 3, 4. For divisibility by 2, the unit digit must be 0, 2, or 4. If the unit digit is 0, the remaining places can be filled in 3! = 6 ways. If the unit digit is 2, the thousands place can be filled in 2 ways (excluding 0), and the remaining places in 2! = 2 ways, giving 2 2 = 4 ways. If the unit digit is 4, similarly, there are 2 2 = 4 ways. Total numbers in Case 1 = 6 + 4 + 4 = 14 Case 2: The excluded digit is 4. The selected digits are 0, 1, 2, 3. For divisibility by 2, the unit digit must be 0 or 2. If the unit digit is 0, the remaining places can be filled in 3! = 6 ways. If the unit digit is 2, the thousands place can be filled in 2 ways (excluding 0), and the remaining places in 2! = 2 ways, giving 2 2 = 4 ways. Total numbers in Case 2 = 6 + 4 = 10 Total favorable outcomes = 14 + 10 = 24 Required probability = 24 96 = 1 4 Answer: 1 4